In a simple harmonic oscillator,at the mean position:

Write the expression of kinetic energy and potential energy of $SHM$ particle $(i)$ as a function of displacement $(ii)$ as a function of time.

$A$ body is performing simple harmonic motion with an amplitude of $10 \, cm$. The velocity of the body is tripled by an air jet when it is at $5 \, cm$ from its mean position. The new amplitude of vibration is $\sqrt{x} \, cm$. The value of $x$ is . . . . . . .

The amplitude of a particle executing $SHM$ is $3\,cm$. The displacement at which its kinetic energy will be $25\%$ more than the potential energy is: $.............cm$.

$A$ linear harmonic oscillator has a total mechanical energy of $300 \ J$. If its potential energy at the mean position is $100 \ J$,find its kinetic energy at $x = +\frac{A}{\sqrt{2}}$. (in $J$)

$A$ particle of mass $m$ is executing simple harmonic motion about its mean position. If $A$ is the amplitude and $T$ is the period of $S$.$H$.$M$.,then the total energy of the particle is